Question

What are the major species present in 0.250$M$ solutions of each of the following acids? Calculate the pH of each of these solutions. a. ${\mathrm{HOC}}_6{\mathrm{H}}_5$ b. HCN

Short answer

For a 0.250 M solution of ${\mathrm{HOC}}_6{\mathrm{H}}_5$, the major species present are the intact acid molecule (${\mathrm{HOC}}_6{\mathrm{H}}_5$), its conjugate base (${\mathrm{OC}}_6{\mathrm{H}}_5^{-}$), and water (${\mathrm{H}}_2\mathrm{O}$) with hydroxide ion (${\mathrm{OH}}^{-}$). Using the Ka value and equilibrium expression, we can calculate the pH of the solution. Similarly, for a 0.250 M solution of HCN, the major species present are the intact acid molecule (HCN), its conjugate base (${\mathrm{CN}}^{-}$), and water (${\mathrm{H}}_2\mathrm{O}$) with hydroxide ion (${\mathrm{OH}}^{-}$). Using the Ka value and equilibrium expression, we can calculate the pH of the solution.

Step-by-step solution

Major Species

For a weak acid ${\mathrm{HOC}}_6{\mathrm{H}}_5$ in solution, we have: 1. The intact acid molecule (${\mathrm{HOC}}_6{\mathrm{H}}_5$) 2. The conjugate base, after releasing a proton: ${\mathrm{OC}}_6{\mathrm{H}}_5^{-}$ 3. Water (${\mathrm{H}}_2\mathrm{O}$) and the hydroxide ion (${\mathrm{OH}}^{-}$) from the auto-ionization of water. Step 2: Determine the equilibrium expression

Equilibrium Expression

The equilibrium expression for the ionization of ${\mathrm{HOC}}_6{\mathrm{H}}_5$ is: $${\mathrm{HOC}}_6{\mathrm{H}}_5\rightleftharpoons {\mathrm{H}}^{+}+{\mathrm{OC}}_6{\mathrm{H}}_5^{-}$$ The equilibrium constant Ka can be expressed as: $$K_a=\frac{[{\mathrm{H}}^{+}][{\mathrm{OC}}_6{\mathrm{H}}_5^{-}]}{[{\mathrm{HOC}}_6{\mathrm{H}}_5]}$$ Step 3: Calculate the pH of 0.250 M ${\mathrm{HOC}}_6{\mathrm{H}}_5$

Calculate pH

Assuming the initial concentration of the acid as 0.250 M and negligible ionization of water, we will solve the Ka expression for [H+] and then use the pH formula: $$\mathrm{pH}=-\log [{\mathrm{H}}^{+}]$$ Using the Ka values for ${\mathrm{HOC}}_6{\mathrm{H}}_5$, we can obtain the pH. #b. Calculation for HCN solution# Step 1: Identify the major species

Major Species

For a weak acid HCN in solution, we have: 1. The intact acid molecule (HCN) 2. The conjugate base, after releasing a proton: ${\mathrm{CN}}^{-}$ 3. Water (${\mathrm{H}}_2\mathrm{O}$) and the hydroxide ion (${\mathrm{OH}}^{-}$) from the auto-ionization of water. Step 2: Determine the equilibrium expression

Equilibrium Expression

The equilibrium expression for the ionization of HCN is: $$\mathrm{HCN}\rightleftharpoons {\mathrm{H}}^{+}+{\mathrm{CN}}^{-}$$ The equilibrium constant Ka can be expressed as: $$K_a=\frac{[{\mathrm{H}}^{+}][{\mathrm{CN}}^{-}]}{[\mathrm{HCN}]}$$ Step 3: Calculate the pH of 0.250 M HCN

Calculate pH

Assuming the initial concentration of the acid as 0.250 M and negligible ionization of water, we will solve the Ka expression for [H+] and then use the pH formula: $$\mathrm{pH}=-\log [{\mathrm{H}}^{+}]$$ Using the Ka values for HCN, we can obtain the pH.

Key concepts

Weak Acids

Weak acids are substances that partially ionize in water. This means that only a small portion of the acid molecules donate their hydrogen ions (protons) to the solution. Because of this partial ionization, weak acids do not completely dissociate into their ions like strong acids do.

Their distinctive characteristics include:

• Lower degree of ionization
• Presence of both unionized acid molecules and ions in the solution
• Equilibrium between the ionized and unionized forms

Examples of weak acids include acetic acid, formic acid, and hydrocyanic acid (HCN). Calculating pH for weak acids tends to be more complex due to the partial ionization involved.

Equilibrium Expression

The concept of equilibrium expression is crucial when dealing with weak acids. The ionization of a weak acid is expressed as an equilibrium between the acid and its ions in water. This balance means that some of the acid molecules donate protons to form ions, while others remain unchanged.

The equilibrium constant for the ionization reaction, represented as $K_a$, indicates how far the reaction goes towards ionization:

• Higher $K_a$ values suggest stronger ionization
• Lower $K_a$ values indicate weaker ionization

For instance, the ionization equilibrium expression for phenol (${\mathrm{HOC}}_6{\mathrm{H}}_5$) is written as: $${\mathrm{HOC}}_6{\mathrm{H}}_5\rightleftharpoons {\mathrm{H}}^{+}+{\mathrm{OC}}_6{\mathrm{H}}_5^{-}$$ This balance allows chemists to calculate concentrations of ions, like $[{\mathrm{H}}^{+}]$, thus enabling pH calculation.

Ionization of Acids

The ionization of acids is the process by which acid molecules release protons into the solution. For weak acids, this process is not complete, leading to a mixture of ions and un-ionized molecules. The extent of ionization is governed by the specific $K_a$ of each acid, which offers insight into how readily an acid gives up protons.

• Weak acids partially release ${\mathrm{H}}^{+}$ in water
• The remaining molecules stay as the original acid

Taking hydrocyanic acid (HCN) as an example, its ionization equation can be represented by: $$\mathrm{HCN}\rightleftharpoons {\mathrm{H}}^{+}+{\mathrm{CN}}^{-}$$This reaction equilibrium is shifted towards intact HCN molecules, reflecting its weak acid nature. Understanding this balance is essential for calculating the pH accurately.

Acid-Base Chemistry

Acid-base chemistry explores the behavior and properties of acids and bases, such as pH, ionization, and equilibrium. In the context of weak acids, it becomes particularly interesting due to their partial ionization. The pH, a measure of acidity or basicity, can be determined by calculating the concentration of ${\mathrm{H}}^{+}$ ions in solution.

• The relationship between ${\mathrm{H}}^{+}$ concentration and pH follows: $\mathrm{pH}=-\log [{\mathrm{H}}^{+}]$
• Weak acids often have pH values higher than their strong acid counterparts at a similar concentration, due to less ionization

Analyzing acids like phenol and HCN, chemists calculate pH by first determining $[{\mathrm{H}}^{+}]$ using the equilibrium expressions derived from $K_a$. This process illustrates the intricate and fascinating interactions within acid-base chemistry.